What is the connection between momentum and energy?

Which of the answers is the correct?

  • A particle can have zero momentum but energy.
  • A particle can have zero energy but momentum.
  • Energy and momentum are always unambiguously connected.

Well, I zero momentum but energy could be potential energy, but that is about it. It must have zero velocity (or zero mass) and therefore have zero kinetic energy.

If it has zero energy, it has zero velocity and therefore zero momentum.

But momentum is a vector and depends on the direction, energy is a scalar and does not care for the direction. So the connection is not that unambiguous after all.

What is the correct answer to this problem?

  • 2
    $\begingroup$ Answer: the question is ill-posed. $\endgroup$ – yohBS Feb 16 '12 at 13:12
  • $\begingroup$ So if you would like to make the best out of it, the third answer would be the most likely to be graded as the correct one? $\endgroup$ – Martin Ueding Feb 16 '12 at 13:56

A is correct, b and c depend upon your framework.

Classical mechanics-Total energy

In classical mechanics, $p=mv$, $E=1/2mv^2+PE$. Potential energy is an arbitrary settable value (you can choose different reference points for zero PE, so energy can be zero regardless of momentum. Momentum can be zero just by the virtue of the particle being at rest. So we get a,b, and not c(since both p and E are independant here)

Classical- kinetic energy

If energy is only referring to kinetic energy, then the answer becomes maybe c. The question itself is quite ambiguous, and the meaning of 'unambiguously conncted' is meta-ambiguous. Are they asking if the connection is ambiguous? If so, then they are unambiguously connected ($|\vec{p}|=\sqrt{2mK}$). On the other hand, they might ask if one can unambiguously derive one from the other. This is false due to the vector thing you mentioned.


Over here, the answers become ac, and its much clearer from this viewpoint. Energy is never zero (Due to $E_{rest}=m_0c^2$ or $E_{tot}=\frac{m_0c^2}{\sqrt{1-v^2/c^2}}$). So b is false. We can still have zero momentum, so a is true. Finally, we have $E^2=m_0^2c^4+c^2p^2$. We still have the meta-ambiguousness, so c may or may not be true.

Summing up

The question has three ambiguosities: the type of energy, the framework, and the meaning of 'unambiguous connection'. Theres also the added confusion of noninertial frames, which I won't go into here.

  • $\begingroup$ Thanks for the analysis. I think the relativity can be ruled out. And since another questions states that multiple answers are correct, but not on this one, I assume it has to be $C$ since we got down to $(A \wedge B) \vee C$, so $C$ is the only single answer that somewhat makes sense. $\endgroup$ – Martin Ueding Feb 16 '12 at 15:00
  • $\begingroup$ You misunderstand me. The kinetic energy/total energy divide depends upon the intention of the examiner. C is quite iffy on its own.. If the question was trying to trip you up, then it probably wants total energy; but if it was a plain, badly worded, but innocent question, then go with kinetic energy. C has its own issues. So its more of $(A\veebar B)\vee (C\vee C')=\monkeyscratchinghead$. $\endgroup$ – Manishearth Feb 16 '12 at 15:12
  • 1
    $\begingroup$ Track down whoever created the question and ask him. Only way to be completely sure =P. $\endgroup$ – Manishearth Feb 16 '12 at 15:13

You need to define what you mean "energy" more carefully.

For example one definition that seems to make sense is to compare momentum and kinetic energy as measured in the same inertial frame, in which case the answers are false, false, true.

However if your framework is special relativity things get more interesting because the following equation always holds:

$$E^2 - c^2p^2 = m_0^2c^4$$

but note that now the energy is the total energy including the rest mass energy as defined by Einstein's famous $e = mc^2$. In this framework you can have energy but no momentum, but you can't have momentum but no energy. So the answers would be true, false, true.

  • $\begingroup$ We can rule out relativity here, so it shoud be $C$ then. $\endgroup$ – Martin Ueding Feb 16 '12 at 15:01

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